0
$\begingroup$

Find the limit without using L'Hopital's rule

$$\lim_{ x \to0 }\frac{x-\sin x}{x^3}=?$$ This was what I did but I would like another solution.

  • 0
    You should say "without L'Hopital" in the title and in the body of the question. – 2017-02-16
  • 0
    Do you want to allow power series? – 2017-02-16
  • 0
    Just use Taylor series – 2017-02-16
  • 3
    You might TeX something from the picture. Not all people understand what the (Arabic) numbers mean. – 2017-02-16
  • 0
    See also : http://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%c3%b4pital-rule-or-series-expansion – 2017-02-16

3 Answers 3

4

If you like an outrageous overkill, from the Weierstrass product $$ \frac{\sin x}{x}=\prod_{n\geq 1}\left(1-\frac{x^2}{n^2\pi^2}\right)\tag{1} $$ it follows that in a neighbourhood of the origin we have: $$ 1-\frac{\zeta(2)}{\pi^2}x^2 \leq \frac{\sin x}{x}\leq \exp\left(-\frac{\zeta(2)}{\pi^2}x^2\right)\tag{2} $$ and the given limit equals $\frac{\zeta(2)}{\pi^2}=\color{red}{\large\frac{1}{6}}$ by squeezing.

2

No Taylor, no L'Hopital: Note that $x-\sin x = \int_0^x (1-\cos t)\, dt.$ Integrating again gives

$$x-\sin x = \int_0^x \int_0^t\sin s \, ds\, dt.$$

We know for small $s$ that $\sin s \sim s.$ So let's use $s$ in place of $\sin s$ above to see what's going on. We get

$$\int_0^x \int_0^t s \, ds\, dt = \int_0^x (t^2/2) dt = x^3/6.$$

Dividing that by $x^3$ gives a limit of $1/6.$ That shows nicely where the answer comes from. All that's left is to make sure using $s$ in place of $\sin s$ above is legitimate.

  • 0
    And how do you show it is legitimate? – 2017-02-16
  • 0
    @MathematicsStudent1122 Give yourself an $\epsilon.$ Then $(1-\epsilon)s < \sin s < (1+\epsilon) s$ for small positive $s.$ Proceed ... – 2017-02-20
1

power series at $$x=0$$ of $$\frac{x-\sin(x)}{x^3}$$ gives $$\frac{1}{6}-\frac{x^2}{120}+\frac{x^4}{540}$$+Terms of the order $$x^6$$ thus our Limit is $$\frac{1}{6}$$